Codeforces Beta Round #89 (Div. 2)D. Caesar's Legions

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本文深入探讨了排序算法与动态规划在数据结构领域的应用，详细阐述了如何利用这些算法解决实际问题，并提供了多个实例进行解析。

Gaius Julius Caesar, a famous general, loved to line up his soldiers. Overall the army had n1 footmen and n2horsemen. Caesar thought that an arrangement is not beautiful if somewhere in the line there are strictly more thatk1 footmen standing successively one after another, or there are strictly more than k2 horsemen standing successively one after another. Find the number of beautiful arrangements of the soldiers.

Note that all n1 + n2 warriors should be present at each arrangement. All footmen are considered indistinguishable among themselves. Similarly, all horsemen are considered indistinguishable among themselves.

Input

The only line contains four space-separated integers n1, n2, k1, k2 (1 ≤ n1, n2 ≤ 100, 1 ≤ k1, k2 ≤ 10) which represent how many footmen and horsemen there are and the largest acceptable number of footmen and horsemen standing in succession, correspondingly.

Output

Print the number of beautiful arrangements of the army modulo 100000000 (108). That is, print the number of such ways to line up the soldiers, that no more than k1 footmen stand successively, and no more than k2horsemen stand successively.

Sample test(s)

input
2 1 1 10

output
1

input
2 3 1 2

output
5

input
2 4 1 1

output
0

Note

Let's mark a footman as 1, and a horseman as 2.

In the first sample the only beautiful line-up is: 121

In the second sample 5 beautiful line-ups exist: 12122, 12212, 21212, 21221, 22121

题意：有n1个步兵，n2个骑兵，将这些士兵进行排队，步兵不能连续排超过k1个，骑兵不能连续排超过k2个，问你排列

的方案数。

思路：dp[i][j][0]表示排列i个步兵，j个骑兵满足步兵要求的方案数：dp[i][j][0]+=dp[i-k][j][1](k表示连续放k个步兵 dp[i][j][1],表示满足骑兵的方法)

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mod 100000000

int dp[105][105][2];
int main()
{
    int n1,n2,k1,k2;
    int i,j,k;
    while(scanf("%d%d%d%d",&n1,&n2,&k1,&k2)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=dp[0][0][1]=1;
        for(i=0;i<=n1;i++)
        {
            for(j=0;j<=n2;j++)
            {
                for(k=1;k<=k1&&k<=i;k++)
                {
                    dp[i][j][0]=(dp[i][j][0]+dp[i-k][j][1])%mod;
                }
                for(k=1;k<=k2&&k<=j;k++)
                {
                    dp[i][j][1]=(dp[i][j][1]+dp[i][j-k][0])%mod;
                }
            }
        }
        printf("%d\n",(dp[n1][n2][0]+dp[n1][n2][1])%mod);
    }
    return 0;
}